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```Section 8-6
Standard Deviation or
Variance
Slide
1
Key Concept
This section introduces methods for testing a
deviation σ or population variance σ 2. The
methods of this section use the chi-square
distribution that was first introduced in
Section 7-5.
Slide
2
Requirements for Testing
Claims About  or  2
1. The sample is a simple random
sample.
2. The population has a normal
distribution. (This is a much stricter
requirement than the requirement of a
normal distribution when testing
Slide
3
Chi-Square Distribution
Test Statistic
2=
n
(n – 1) s 2
2
= sample size
s 2 = sample variance
2 = population variance
(given in null hypothesis)
Slide
4
P-Values and Critical Values for
Chi-Square Distribution
 Use Table A-4.
 The degrees of freedom = n –1.
Slide
5
Properties of Chi-Square
Distribution
 All values of  2 are nonnegative, and the
distribution is not symmetric
(see Figure 8-13, following).
 There is a different distribution for each
number of degrees of freedom
(see Figure 8-14, following).
 The critical values are found in Table A-4
using n – 1 degrees of freedom.
Slide
6
Properties of Chi-Square
Distribution - cont
Properties of the Chi-Square
Distribution
Chi-Square Distribution for 10
and 20 Degrees of Freedom
There is a different distribution for each
number of degrees of freedom.
Figure 8-13
Figure 8-14
Slide
7
Example: For a simple random sample of adults, IQ scores are
normally distributed with a mean of 100 and a standard deviation
of 15. A simple random sample of 13 statistics professors yields a
standard deviation of s = 7.2. Assume that IQ scores of statistics
professors are normally distributed and use a 0.05 significance
level to test the claim that  = 15.
H0:  = 15
H1:   15
 = 0.05
n = 13
s = 7.2
 =
2
(n – 1)s2
2
2
(13
–
1)(7.2)
=
= 2.765
152
Slide
8
Example: For a simple random sample of adults, IQ scores are
normally distributed with a mean of 100 and a standard deviation
of 15. A simple random sample of 13 statistics professors yields a
standard deviation of s = 7.2. Assume that IQ scores of statistics
professors are normally distributed and use a 0.05 significance
level to test the claim that  = 15.
H0:  = 15
H1:   15
 = 0.05
n = 13
s = 7.2
 2 = 2.765
Slide
9
Example: For a simple random sample of adults, IQ scores are
normally distributed with a mean of 100 and a standard deviation
of 15. A simple random sample of 13 statistics professors yields a
standard deviation of s = 7.2. Assume that IQ scores of statistics
professors are normally distributed and use a 0.05 significance
level to test the claim that  = 15.
H0:  = 15
H1:   15
 = 0.05
n = 13
s = 7.2
 2 = 2.765
The critical values of 4.404 and 23.337 are
found in Table A-4, in the 12th row (degrees
of freedom = n – 1) in the column
corresponding to 0.975 and 0.025.
Slide
10
Example: For a simple random sample of adults, IQ scores are
normally distributed with a mean of 100 and a standard deviation
of 15. A simple random sample of 13 statistics professors yields a
standard deviation of s = 7.2. Assume that IQ scores of statistics
professors are normally distributed and use a 0.05 significance
level to test the claim that  = 15.
H0:  = 15
H1:   15
 = 0.05
n = 13
s = 7.2
 2 = 2.765
Because the test statistic is in the critical
region, we reject the null hypothesis. There
is sufficient evidence to warrant rejection of
the claim that the standard deviation is equal
to 15.
Slide
11
Recap
In this section we have discussed:
 Tests for claims about standard deviation
and variance.
 Test statistic.
 Chi-square distribution.
 Critical values.
Slide
12
```
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